Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100000 * 4;
const long long mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const int DEG = 20;
struct Edge {
  int to, next;
} e[maxn * 2];
int head[maxn], tot;
vector<int> dp[maxn][DEG], ins[maxn];
void addedge(int u, int v) {
  e[tot] = {v, head[u]};
  head[u] = tot++;
}
void init() {
  tot = 0;
  memset(head, -1, sizeof(head));
}
int fa[maxn][DEG];
int deg[maxn];
vector<int> cal(vector<int> &v1, vector<int> &v2) {
  vector<int> dst;
  dst.clear();
  std::merge(v1.begin(), v1.end(), v2.begin(), v2.end(),
             std::back_inserter(dst));
  vector<int>::iterator iter = unique(dst.begin(), dst.end());
  dst.erase(iter, dst.end());
  while (dst.size() > 10) {
    dst.pop_back();
  }
  return dst;
}
void BFS(int root) {
  queue<int> que;
  while (!que.empty()) que.pop();
  deg[root] = 0;
  fa[root][0] = 0;
  que.push(root);
  while (!que.empty()) {
    int v = que.front();
    que.pop();
    dp[v][0] = cal(ins[v], ins[fa[v][0]]);
    for (int i = 1; i < DEG; i++) {
      fa[v][i] = fa[fa[v][i - 1]][i - 1];
      dp[v][i] = cal(dp[v][i - 1], dp[fa[v][i - 1]][i - 1]);
    }
    for (int i = head[v]; i != -1; i = e[i].next) {
      int to = e[i].to;
      if (to == fa[v][0]) continue;
      deg[to] = deg[v] + 1;
      fa[to][0] = v;
      que.push(to);
    }
  }
}
vector<int> LCA(int u, int v) {
  vector<int> res;
  res.clear();
  res = cal(ins[u], ins[v]);
  if (deg[u] > deg[v]) {
    swap(u, v);
  }
  int hu = deg[u], hv = deg[v];
  int tu = u, tv = v;
  int flag = 0;
  for (int det = hv - hu, i = 0; det; det >>= 1, i++) {
    if (fa[v][i] != 0 && det & 1) {
      res = cal(res, dp[tv][i]);
      tv = fa[tv][i];
      flag = 1;
    }
  }
  if (tu == tv) {
    if (!flag) res = cal(ins[tu], res);
    return res;
  }
  for (int i = DEG - 1; i >= 0; i--) {
    if (fa[tu][i] == fa[tv][i]) {
      continue;
    }
    res = cal(res, dp[tv][i]);
    res = cal(res, dp[tu][i]);
    tu = fa[tu][i];
    tv = fa[tv][i];
  }
  res = cal(res, ins[fa[tv][0]]);
  return res;
}
int n, m, q;
int main() {
  if (fopen("/home/lianyi/123.txt", "r") != NULL) {
    freopen("/home/lianyi/123.txt", "r", stdin);
  } else {
    std::ios::sync_with_stdio(false);
    cin.tie(0);
  }
  while (cin >> n >> m >> q) {
    init();
    int u, v;
    for (int i = 1; i < n; i++) {
      cin >> u >> v;
      addedge(u, v);
      addedge(v, u);
    }
    for (int i = 1; i <= n; i++) {
      ins[i].clear();
      for (int j = 0; j < DEG; j++) {
        dp[i][j].clear();
      }
    }
    int ci;
    for (int i = 1; i <= m; i++) {
      cin >> ci;
      if (ins[ci].size() < 10) {
        ins[ci].push_back(i);
      }
    }
    BFS(1);
    int a;
    for (int i = 1; i <= q; i++) {
      cin >> u >> v >> a;
      vector<int> res = LCA(u, v);
      while (res.size() > a) {
        res.pop_back();
      }
      cout << res.size();
      if (res.size() == 0) {
        cout << '\n';
      } else {
        cout << ' ';
      }
      for (int i = 0; i < res.size(); i++) {
        cout << res[i];
        if (i == res.size() - 1) {
          cout << '\n';
        } else {
          cout << ' ';
        }
      }
    }
  }
  return 0;
}